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\title{\vspace{-4cm}\textbf{河北师范大学数学分析真题}}
\author{宁鑫雨}
\date{\today}
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\begin{document}
\date{}
\section*{2012年数学分析}
\begin{problem}[本题满分60分，每小题10分]计算下列各题:\\
    1、求极限$\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2-2}}-\cdots-\frac{1}{\sqrt{n^2-n}}\right)$.\\
    2、计算积分$I=\int_{0}^{4}e^{\sqrt{x}}\ d x$.\\
    3、求级数$\sum_{n=2}^{\infty}\frac{1}{(n^{2}-1)3^{n}}$的和.\\
    4、计算二重积分$I=\iint_D|x-y|\ d x \ d y$,其中$D$为半径为$R$的圆域$x^{2}+y^{2}=R^{2}$的上半部分.\\
    5、计算$\begin{aligned}I=\iint_{\sum}(8y+1)x\ d y \ d z+2(1-y^{2})\ d z \ d x-4yz\ d x \ d y\end{aligned}$.\\
    其中$\sum$是曲线${L}:\begin{cases}z=\sqrt{y-1}\\x=0\end{cases}\quad(1\leq y\leq3)$绕$y$轴旋转所成的曲面的外侧.\\
    6、已知曲线$y=f(x)$在$x=1$处的切线方程为：$y-2x+2=0$,求极限:$\lim_{x\to0}\frac{\int_{0}^{x}e^{t}f(1+e^{x}-e^{t})dt}{1-\cos x}.$
\end{problem}
\begin{solution}[本题60分,每题10分]\\
    1)$\lim_{n\to\infty}\left[\frac{1}{\sqrt{n^{2}-1}}-\left(\frac{1}{\sqrt{n^{2}-2}}+\cdots+\frac{1}{\sqrt{n^{2}+n}}\right)\right]$\\
    $\frac{n}{\sqrt{n^{2}-2}}<\frac{1}{\sqrt{n^{2}-2}}+\cdots+\frac{1}{\sqrt{n^{2}-n}}<\frac{n}{\sqrt{n^{2}-n}}$\\
    $\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^{2}-2}}=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{1-\frac{2}{n}}}=1$\\
    $\lim_{n\to\infty}\frac{n}{\sqrt{n^{2}-n}}=\lim_{n\to\infty}\frac{1}{\sqrt{1-\frac{1}{n}}}=1$\\
    $\therefore\lim_{n\to\infty}(\frac{1}{\sqrt{n^{2}-2}}+\cdots+\frac{1}{\sqrt{n^{2}-n}})=1$\\
    又$\lim_{n\to\infty}\frac{1}{\sqrt{n^{2}-1}}=0$\\
    $
    \begin{aligned}
        \therefore &\lim_{n\to\infty}(\frac{1}{\sqrt{n^{2}-1}}-\frac{1}{\sqrt{n^{2}-2}}-\cdots-\frac{1}{\sqrt{n^{2}-n}})\\
        &=\lim_{n\to\infty}\left[\frac{1}{\sqrt{n^{2}-1}}-\left(\frac{1}{\sqrt{n^{2}-2}}+\cdots+\frac{1}{\sqrt{n^{2}+n}}\right)\right] \\
        &=-1
    \end{aligned}$\\
    2)令$t=\sqrt{x}$\\
    $\begin{aligned}
        &I=\int_{0}^{4}e^{\sqrt{x}}\ d x=2\int_{0}^{2}te^{t}\ d t=2\int_{0}^{2}tde^{t} \\
        &=2[te^{t}\big|_{0}^{2}-\int_{0}^{2}e^{t}dt]=2[2e^{2}-e^{t}\big|_{0}^{1}]=2(e^{2}+1)
    \end{aligned}$\\
    3)\\
    $\begin{aligned}
        &\sum_{n=2}^{\infty}\frac{1}{(n^{2}-1)3^{n}} \\
        &=\sum_{n=2}^{\infty}\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\cdot\frac{1}{3^{n}} \\
        &=\frac{1}{2}\sum_{n=2}^{\infty}\frac{1}{n-1}\cdot\frac{1}{3^{n}}-\frac{1}{2}\sum_{n=2}^{\infty}\frac{1}{n+1}\cdot\frac{1}{3^{n}} \\
        &=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}\cdot\frac{1}{3^{n+1}}-\frac{1}{2}\sum_{n=2}^{\infty}\frac{1}{n+1}-\frac{1}{3^{n}}
    \end{aligned}$\\
    令$f(x)=\sum_{n=1}^{\infty}\frac{1}{n}\cdot x^{n+1}=x\cdot\sum_{n=1}^{\infty}\frac{1}{n}\cdot x^{n}$\\
    令$g(x)=\sum_{n=1}^{\infty}\frac{1}{n}x^{n}$\\
    $g^{\prime}(x)=(\sum_{n=1}^{\infty}\frac{1}{n}x^{n})^{\prime}=\sum_{n=1}^{\infty}x^{n-1}=\frac{1}{1-x}$\\
    $\therefore g(x)=\int_{0}^{x}\frac{1}{1-t}\ d t=-\ln(1-t)\big|_{0}^{x}=-\ln(1-x)$\\
    $\therefore f(x)=-x\ln(1-x)$\\
    $f(\frac{1}{3})=-\frac{1}{3}\ln\left(\frac{2}{3}\right)$\\
    令$m(x)=\sum_{n=2}^{\infty}\frac{1}{n+1}x^{n}$\\
    $N(x)=\sum_{n=2}^{\infty}\frac{1}{n+1}x^{n+1}=m(x)\cdot x$\\
    $N^{\prime}(x)=\left(\sum_{n=2}^{\infty}\frac{1}{n+1}x^{n+1}\right)^{\prime}=\sum_{n=2}^{\infty}x^{n}=\frac{x^{2}}{1-x}.$\\
    $\begin{aligned}
        &\therefore N(x)=\int_{0}^{\infty}\frac{t^{2}}{1-t}\ d t=\int_{0}^{x}\frac{t^{2}-1+1}{1-t}\ d t \\
        &=\int_{0}^{x}\frac{t^{2}-1}{1-t}\ d t+\int_{0}^{x}\frac{1}{1-t}\ d t \\
        &=-\int_{0}^{x}(t+1)\ d t-\int_{0}^{x}\frac{1}{1-t}\ d (1-t) \\
        &=-\frac{1}{2}x^{2}-x-\ln(1-x)
    \end{aligned}$\\
    $\therefore m(x)=\frac{N(x)}{x}=-\frac{1}{2}x-1-\frac{ln(1-x)}{x}$\\
    $m(\frac{1}{3})=-\frac{1}{6}-1-\frac{ln\frac{2}{3}}{\frac{1}{3}}$\\
    $\begin{gathered}
        \therefore\sum_{n=2}^{\infty}\frac{1}{(n^{2}-1)3^{n}}=\frac{1}{2}(-\frac{1}{3}ln\frac{2}{3})+\frac{1}{2}(\frac{7}{6}+3\ln\frac{2}{3}) \\
        =\frac{7}{12}+\frac{4}{3}\ln\frac{2}{3} 
    \end{gathered}$\\
    4)$\begin{aligned}
        &I=\iint|x-y|\ d x \ d y\\
        &=\iint_{D_1}(x-y)\ d x \ d y+\iint_{D_2}(y-x)\ d x \ d y\\
        &=\int_{0}^{\frac{\pi}{4}}\ d \theta\int_{0}^{R}(r\cos\theta-r\sin \theta)r\ d r+\int_{\frac{\pi}{4}}^{\pi}\ d \theta\int_{0}^{R}(r\sin\theta-r\cos\theta)r\ d r\\
        &=\frac{\sqrt{2}-1}{3}R^{3}+\frac{\sqrt{2}+1}{3}R^{3}=\frac{2\sqrt{2}}{3}R^{3}
    \end{aligned}$\\
    5)略\\
    6)易得$f(1)=0,f^{\prime\prime}(1) =2$\\
    令$u=1+e^{x}-e^{t},\ d u=-e^{t}\ d t$\\
   $
   \begin{aligned}
    \therefore&\lim_{x\to0}\frac{\int_{0}^{x}e^{t}f(1+e^{x}-e^{t})\ d t}{1-\cos x}=\lim_{x\to0}\frac{\int_{1}^{e^{x}}f(u)\ d u}{1-\cos x}\\
        &=\lim_{x\to0}\frac{e^{x}f(e^{x})}{\sin x}=\lim_{x\to0}\frac{e^{x}(f(e^{x})+f(e^{x})\cdot e^{x})}{\cos x}\\
        &=f(1)+f^{\prime\prime}(1)\\
        &=2
    \end{aligned}$
\end{solution}


\begin{problem}[本题满分15分]\\
    设$f(x)$在$[a,b]$二阶可导，且$f(b)=0$,令$g(x)=(x-a)^2f(x)$,证明方程$g^{\prime\prime}(x)=0$在$(a,b)$上有解.
\end{problem}
\begin{solution}[本题15分]\\
    证明：\\
    $g(x)=(x-a)^{2}f(x)$,$\therefore g(a)=0$\\
    又$\because f(b)=0$,$\therefore g(b)=0$\\
   $\because $ $f(x)$ 在 $[a,b]$ 二阶可导,\\
   $\therefore g(x)$ 在 $[a,b]$ 连续可导\\
   $\therefore\exists\xi\in(a,b)$,使得$g^{\prime}(\xi)=0$\\
   $g^{\prime}(x)=2(x-a)f(x)+(x-a)^{2}f^{\prime}(x)$在$[a,b]$连续,$(a,b)$可导\\
   $g^{\prime}(a)=0$\\
   $\therefore\exists \eta \in (a,\xi)\subset (a,b)$,使得$g^{\prime}(\eta)=0$\\
   $\therefore g^{\prime\prime}(x)=0$在$(a,b)$上有解.
\end{solution}


\begin{problem}[本题满分15分]\\
    设$f(x)={x}e^{-x^{2}}\int_{0}^{x}e^{t^{2}}\ d t$\\
    1、求$\lim_{x\rightarrow+\infty}f(x)$,并证明$f(x)$在$[0,+\infty)$有界；\\
    2、证明$x\int_{0}^{\pi}e^{t^{2}}\ d t\leq e^{x^{2}}-1,x\in[0,+\infty)$.
\end{problem}
\begin{solution}[本题15分]\\
   1)设$\int_{0}^{x}e^{t^{2}}\ d t=g(x)$\\
   $\begin{aligned}
    &\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}xe^{-x^{2}}g(x)\\
    &=\lim_{x\to+\infty}\frac{x\cdot g(x)}{e^{x^{2}}}\\
    &=\lim_{x\to+\infty}\frac{g(w)+pe^{x^{2}}}{e^{x^{2}}-x}\\
    &=\lim_{x\to+\infty}\frac{e^{x^{2}}+e^{x^{2}}+x\cdot e^{x^{2}}\cdot 2x}{e^{x^{2}} \cdot (2x)^{2}+e^{x^{2}}\cdot2}\\
    &=\lim_{x\to+\infty}\frac{1+x^{2}}{2x^{2}+1}=\frac{1}{2}
\end{aligned}$\\
$\therefore\lim_{x\to\infty}f(x)=\frac{1}{2}$\\
$\therefore \varepsilon>0,\exists M,$当$x>M$时,$|f(x)-\frac{1}{2}|<\varepsilon $\\
$\therefore\frac{1}{2}-\varepsilon<f(x)<\frac{1}{2}+\varepsilon$\\
$\therefore$$f(x)$有界\\
2)设$F(x)=x\int_{0}^{x}e^{t^{2}}\ d t-e^{x^{2}}-1$\\
$F^{\prime}(x)=\int_{0}^{x}e^{t^{2}}\ d t+x\cdot e^{x^{2}}-e^{x^{2}}\cdot 2x=\int_{0}^{x}e^{t^{2}}\ d t-xe^{x^{2}}$
$F^{\prime\prime}(x)=e^{x^{2}}-e^{x^{2}}-x\cdot e^{x^{2}}\cdot2x=-2x^{2}e^{x^{2}}\leq0$\\
$\therefore$$F^{\prime}(x)$单减,$F^{\prime}(0)=0$,$F^{\prime}(x)\leq0$\\
$\therefore$$F(x)$单减,$F(0)=0$,$F(x)\leq0$\\
$\therefore x\int_{0}^{x}e^{t^{2}}\ d t\leq e^{x^{2}}-1$
\end{solution}

    
\begin{problem}[本题满分15分]\\
    证明级数$\sum_{n=1}^{\infty}x^{2}e^{-nx}$在$(0,+\infty)$一致收敛；并说明其和函数$f(x)$的连续性
\end{problem}
\begin{solution}[本题15分]\\
    证明:\\
    $g(x)=x^{2}e^{-nx}$\\
    $\begin{aligned}
        g^{\prime}(x)
        &=2x\cdot e^{-nx}+x^2\cdot e^{-nx}\cdot-n\\
        &=(2-nx)xe^{-nx}
    \end{aligned}$\\
    令$g^{\prime}(x)=0$,$x=\frac{2}{n}$\\
    $\therefore$$g(x)$在$x=\frac{2}{n}$处取最大值\\
    $g(\frac{2}{n})=\frac{4}{n^{2}}\cdot e^{-2}=\frac{4}{n^{2}e^{2}}$\\
    $\therefore|g(x)|=|x^{2}e^{-nx}|\leq\frac{4}{n^{2}e^{2}}$\\
    $\because\sum_{n=1}^{\infty}\frac{4}{n^{2}e^{2}}$收敛,\\
    $\therefore$由判别法知$\sum_{n=1}^{\infty}x^{2}e^{-nx}$一致收敛\\
    又$\because$级数每一项均连续\\
    $\therefore$由连续性定理知其和函数连续
\end{solution}
    

\begin{problem}[本题满分10分]\\
    设$f(x)$在有限区间$(a,b)$一致连续，证明$[f(x)]^{2}$在$(a,b)$一致连续。
\end{problem}
\begin{solution}[本题10分]\\
   证明:\\
   $\because f(x)$在有限区间$(a,b)$一致连续\\
   $\therefore$$\forall \varepsilon>0,\exists \delta>0$,当$x_{1},x_{2}\in (a,b)$,$|x_{1}-x_{2}|<\delta$时,$|f(x_1)-f(x_2)|<\varepsilon$ \\
   设$F(x)=\begin{cases}f(a+0)&x=a\\f(x)&x\in(a,b)\\f(b-0)&x=b\end{cases}$\\
   显然$F(x)$在$[a,b]$上连续$\therefore f(x)$在$[a,b]$上有界\\
   $\therefore f(x)$在$[a,b]$有界，$\therefore f(x)$在$(a,b)$有界，设$|f(x)|<M$\\
   $\begin{aligned}\left|\left[f(x_{1})\right]^{2}-\left[f(x_{2})\right]^{2}\right|&=\left[f(x_{1})+f(x_{2})\right]\left[f(x_{1})-f(x_{2})\right]\\&<\xi\cdot2M\end{aligned}$\\
   $\therefore$$[f(x)]^{2}$在$(a,b)$一致连续
\end{solution}
    

\begin{problem}[本题满分10分]\\
    设函数$f(x)$在$[a,b]$非负可积，在点$x_{0}\in(a,b)$连续且$f(x_{0})>0.$.证明：存在$c$使得$\int_{a}^{b}f(x)dx>c$.
\end{problem}
\begin{solution}[本题10分]\\
    $f(x)\geq0$$f(x)$在$x_{0}$连续,且$f(x_{0})>0$\\
    $\therefore \exists U(x_{0})$,当$x\in U(x_{0}),f(x)>\frac{1}{2}f(x_{0})$\\
    $\begin{aligned}
        \int_{a}^{b}f(x)\ d x&=\int_{a}^{x-\delta}f(x)\ d x+\int_{x_{0}-\delta}^{x+\delta}f(x)\ d x+\int_{x_{0}+\delta}^{b}f(x)\ d x\\
        &\geq\int_{x_{0}-\delta}^{x_{0}+\delta}f(x)\ d x>\int_{x_{0}-\delta}^{x_{0}+\delta}\frac{1}{2}f(x_{0})\ d x\\
        &=f(x_{0})\cdot\delta
    \end{aligned}$\\
    令$c=f(x_{0})\cdot\delta$\\
   $\therefore$ $\int_{a}^{b}f(x)dx>c$
\end{solution}

\begin{problem}[本题满分15分]\\
    定义黎曼函数：$R(x)=\frac{1}{p},x=\frac{q}{p}\quad , p\geq1, p,q$互质；$R(x)=0,x$为无理数.\\
    1、证明$R(x)$在$[0,1]$上可积；\\
    2、证明$R(x)$在$[0,1]$上无原函数；\\
    3、求$\int_0^1R(x)dx$.
\end{problem}
\begin{solution}[本题15分]
    证明:\\
    1)任给$\varepsilon>0$,在$[0,1]$上使得$\frac{1}{p}>\frac{\varepsilon}{2}$的有理点$\frac{q}{p}$只有有限个，设它们分别为$r_{1},\ldots r_{k},$\\
    现为$[0,1]$作分割$T={\Delta_{1},\Delta_{2},\cdots\Delta_{n}}$,使得$\|T\|<\frac{\varepsilon}{2k}$.\\
    并把$T$中所有小区间分为$\{\Delta i^{\prime}|{i=1,2,\ldots,m}\}$和$\left\{\Delta i^{\prime\prime}\mid i=1,2,\cdots,n-m\right\}$两类\\
    其中${\Delta i^{\prime}}$为含有$r_{1},\ldots r_{k},$中点的小区间,这类小区间的个数$m\leq 2k$,而${\Delta i^{\prime\prime}}$为不含有$r_{1},\ldots r_{k},$中点的小区间\\
    由于$f(x)$在${\Delta i^{\prime}}$ 上的振幅 $w_i^{\prime}\leq\frac{1}{2}$,\\
    $\therefore\sum_{i=1}^{m}w_{i}^{\prime}\Delta x_{i}^{\prime}\leq\frac{1}{2}\sum_{i=1}^{m}\Delta x_{i}^{\prime}\leq\frac{1}{2}\cdot2k\cdot\|T\|<\frac{\varepsilon}{2}$\\
    $\therefore\sum_{i=1}^{n}w_{i}\Delta x_{i}=\sum_{i=1}^{m}w_{i}^{\prime}\Delta x_{i}^{\prime}+\sum_{i=1}^{n-m}w_{i}^{\prime\prime}\Delta x_{i}^{\prime\prime}<\varepsilon$\\
    $\therefore f$ 在$[0,1]$上可积\\
    2)设$R(x)$存在原函数$F(x)$,即$F^{\prime}(x)=R(x)$\\
    $F^{\prime}(0)=0,F^{\prime}(\frac{1}{2})=\frac{1}{2}$\\
    $\therefore$由导函数介值定理,$R(x)$可取得$(0,\frac{1}{2})$上任意值\\
    $\therefore$矛盾\\
    $\therefore R(x)$在$[0,1]$上无原函数\\
    3)$\because f$在$[0,1]$可积\\
    取$[0,1]$上任意分割$T$,取$\xi_{i}$均为$[0,1]$上的无理点\\
    $\therefore\int_{0}^{1}f(x)\ d x=\lim_{\|T\|\to0}\sum_{i=1}^{n}f(\xi)\cdot\Delta x_{i}=0$\\
    $\therefore\int_{0}^{1}f(x)\ d x=0$\\
\end{solution}


\begin{problem}[本题满分10分]\\
    设$\{f_{n}\}$为$[a,b]$上的连续函数列，且$f_{1}\geq f_{2}\geq\cdots\cdots$,又$\lim_{n\to\infty}f_{n}\left(x\right)=f\left(x\right)$在$[a,b]$上处处存在，证明$f(x)$在$[a,b]$上有最大值。
\end{problem}    
\begin{solution}[本题10分]
   证明:\\
   $\because$$\{f_{n}\}$上$[a,b]$连续,$\therefore f_{1}(x)$在$[a,b]$连续\\
   $\therefore f_{1}(x)$有界\\
   $\therefore m_{0},x \in[a,b],f_{1}(x)\leq m_{0}$\\
   又$\because f_{1}\geq f_{2}\geq\cdots $ $\therefore \{f_{n}\}$单减\\
   又$\because\lim_{n\to\infty}f_{n}(x)=f(x)$\\
   $\therefore f(x)\leq f_{1}(x)\leq m_{0}$, $\therefore f(x)$有上界\\
   $\therefore f(x)$有上确界,设$M=\max_{x\in[a,b]}f(x)$\\
   $\therefore \forall k,\exists x_{k}\in[a,b],f(x_{k})>M-\frac{1}{k}$\\
   $\because [a,b]$为闭区间,$\therefore$一定存在一个收敛的子列\\
   设为$\{x_{k}\}\subset[a,b],$设$\lim_{k\to\infty}x_{k}=x_{0}\in[a,b]$\\
   $\begin{aligned}
    \therefore f_{n}(x)
    &=f_{n}\left(\lim_{k\to\infty}x_{k}\right)=\lim_{k\to\infty}f_{n}(x_{k})\\
    &\geq\lim_{k\to\infty}f(x_{k})=\lim_{k\to\infty}(M-\frac{1}{k})=M
   \end{aligned}$
   $\therefore\lim_{n\to\infty}f_{n}(x_{0})=f(x_{0})\geq M$\\
   又$\because M$为$f(x)$的上确界,$\therefore f(x_{0})\leq M$\\
   $\therefore f(x_{0})=M$\\
   $\therefore f(x)$在$[a,b]$上有最大值
\end{solution}
\end{document}